Which of the following points lies on the line 2x+3y=5?

Mathematics

1 answer:

mash [69]2 years ago

6 0

Answer:

B(2,1). And C(-2,3)

Step-by-step explanation:

Tbh, you can just plug the numbers into the original equation :2x+3y=5

A.2x2+3x3=13 ≠5

B.2x2+3x1=5=5

C.-2x2+3x3=5=5

D.-2x2+3x0=-4≠5

So the answer is B and C lies on the line

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astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.